Hello,

I'm in minor confusion over slopes involving perpendicular lines.

Consider a rectangular co-ordinate system. The scales I'm using are such that ten small squares on my graph paper represent 1 unit on the x-axis and 2 units on the y-axis. I've drawn a straight line which passes through when . I've now drawn another straight line where both lines intersect at x=1 and are perpendicular (corner-to-corner for each square on the graph paper for both straight lines).

Now I define the equation for the second straight line choosing arbitrary points:

For the point of intersection we solve simultaneously, from which yields x=1. The solution for both equations holds.

Further, two straight lines are perpendicular - if and only if the product of their slopes equates to

Therefore,

Delving deeper, one now remembers such that if you 'somehow' embrace the former line and rotate it through an angle of then you have gone up x units and have subsequently moved towards the left by way of y units across.

We now define the equation for this perpendicular line from which is

Therefore,

I now explain my confusion. This is an absolute load of horse radish. A perpendicular line can only have a negative reciprocal for a slope if one inverts the axes. In other words, the x-axis becomes the y-axis and on the converse the y-axis becomes the x-axis.

What this tells me is that these lines cannot be constructed in an identical rectangular co-ordinate system in order to have the product of their slopes equate to (-1).

For example, if I plot a line by means of subbin' in x values for using the former co-ordinate system I used to construct they're not perpendicular.

What am I missing?

Thank you for your attention.